/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 62 A suspension bridge is \(60.0 \m... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 62

A suspension bridge is \(60.0 \mathrm{m}\) above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistance. At the location of the bridge \(g\) has been measured to be $9.83 \mathrm{m} / \mathrm{s}^{2} .$ (a) If you drop the stone, how long does it take for it to fall to the base of the gorge? (b) If you throw the stone straight down with a speed of \(20.0 \mathrm{m} / \mathrm{s},\) how long before it hits the ground? (c) If you throw the stone with a velocity of \(20.0 \mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal, how far from the point directly below the bridge will it hit the level ground?

Short Answer

Expert verified
Question: Calculate the time it takes for a stone dropped from a suspension bridge, thrown straight down with a speed of 20.0 m/s, and thrown with a velocity of 20.0 m/s at 30掳 above the horizontal, to hit the ground. Also, find the distance from the point directly below the bridge that the stone would hit when thrown at an angle. Answer: When dropped, it takes approximately 3.49 s for the stone to hit the ground. When thrown straight down with a speed of 20.0 m/s, it takes approximately 2.42 s for the stone to hit the ground. When thrown with a velocity of 20.0 m/s at 30掳 above the horizontal, it takes approximately 2.90 s for the stone to hit the ground and lands approximately 50.23 m from the point directly below the bridge.

Step by step solution

01

Case (a) - Free Fall

In this case, the stone is simply dropped from the bridge. We can use the following equation of motion to find the time it takes for the stone to fall to the base of the gorge: \(h = \frac{1}{2}gt^2\) where \(h = 60.0\) m, \(g=9.83\ \mathrm{m/s^2}\), and \(t\) is the time it takes for the stone to fall. Solving for \(t\): \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(60.0)}{9.83}} \approx 3.49 \ \mathrm{s}\) So, the time it takes for the stone to fall is \(3.49 \ \mathrm{s}\).
02

Case (b) - Thrown Straight Down

In this case, the stone is thrown straight down with a speed of \(20.0 \mathrm{m/s}\). Using the following equation of motion: \(h = v_0t + \frac{1}{2}gt^2\) where \(h = 60.0\) m, \(v_0= 20.0 \ \mathrm{m/s}\), \(g=9.83\ \mathrm{m/s^2}\), and \(t\) is the time it takes for the stone to fall. Solving for \(t\): \(t= \frac{-v_0 + \sqrt{v_0^2+2gh}}{g} = \frac{-20 +\sqrt{(20)^2+2(9.83)(60)}}{9.83} \approx 2.42\ \mathrm{s}\) So, when the stone is thrown straight down with a speed of \(20.0 \mathrm{m/s}\), it takes \(2.42\ \mathrm{s}\) for it to hit the ground.
03

Case (c) - Thrown At An Angle

In this case, the stone is thrown with a velocity of \(20.0 \mathrm{m/s}\) at \(30.0^{\circ}\) above the horizontal. First, we need to find the vertical and horizontal components of the velocity: \(v_0x = v_0 \cos(30^\circ) = 20\ \mathrm{m/s} \times \cos(30^\circ) \approx 17.32\ \mathrm{m/s}\) \(v_0y = v_0 \sin(30^\circ) = 20\ \mathrm{m/s} \times \sin(30^\circ) = 10.0\ \mathrm{m/s}\) Now, we'll find the time it takes for the stone to hit the ground using the equation of motion: \(h = v_0yt + \frac{1}{2}gt^2\) Solving for \(t\): \(t= \frac{-v_0y + \sqrt{v_0y^2+2gh}}{g} = \frac{-10 +\sqrt{(10)^2+2(9.83)(60)}}{9.83} \approx 2.90\ \mathrm{s}\) Finally, we'll find the distance from the point directly below the bridge that the stone hits the ground using the following equation: \(x = v_0xt\) \(x = (17.32\ \mathrm{m/s})(2.90\ \mathrm{s}) \approx 50.23\ \mathrm{m}\) So, when the stone is thrown with a velocity of \(20.0\ \mathrm{m/s}\) at \(30.0^\circ\) above the horizontal, it will hit the ground approximately \(50.23\ \mathrm{m}\) from the point directly below the bridge.

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